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Maslowich
3 years ago
9

PLS HELP!I will give brainliest! When producing hydrogen iodide, the energy of the reactants is 581 kJ/mol, and the energy of th

e products is 590 kJ/mol. The equation is shown.
H2 + I2 → 2HI

What is the total energy of the reaction? Is this an endothermic or exothermic reaction?
A. –9 kJ/mol, exothermic
B.9 kJ/mol, exothermic
C. 9 kJ/mol, endothermic
D. –9 kJ/mol, endothermic
Chemistry
1 answer:
Veronika [31]3 years ago
4 0

Answer:

Here, we are required to determine the total energy of the reaction and determine if the reaction is an endothermic or exothermic reaction.

The correct answer is option C.

First, we need to determine the energy of the reaction.

The energy of the reaction is the change in enthalpy between the product and reactants.

Change of Enthalpy,

Hreaction = Hproduct - Hreactant.

Therefore, for the reaction above, the change in enthalpy is:

Hreaction = 590kJ/mol - 581kJ/mol.

Hreaction = 9kJ/mol.

Hence, since the reaction has an enthalpy change of 9kJ/mol, the reaction is endothermic (i.e energy is absorbed).

Explanation:

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Estimate the surface-to-volume ratio of a C60 fullerene by treating the molecule as a hollow sphere and using 77pm for the atomi
AleksAgata [21]

Answer:

The surface-to-volume ratio of a C-60 fullerene is 3:77.

Explanation:

Surface area of sphere = S=4\pi r^2

Volume of the sphere = V=\frac{4}{3}\pi r^3

where : r  = radius of the sphere

Radius of the C-60 fullerene sphere = r = 77 pm

Surface area of the C-60 fullerene = S=4\pi (77 pm)^2...[1]

Volume area of the C-60 fullerene = V=\frac{4}{3}\pi (77 pm)^3..[2]

Dividing [1] by [2]:

\frac{S}{V}=\frac{4\pi (77 pm)^2}{\frac{4}{3}\pi (77 pm)^3}

=\frac{3}{77}

The surface-to-volume ratio of a C-60 fullerene is 3:77.

7 0
3 years ago
What conclusions can you draw about how temperature and salinity affect the flow of water? Write an evidence-based claim.
S_A_V [24]

Temperature and salinity changes the density of water.

<h3>Effect of Temperature and salinity on water</h3>

Temperature and salinity directly affect density of water. Water that low temperature is more denser than water that has high temperature while on the other hand, Freshwater which has no salt is less denser than seawater which has more salt concentration so we can conclude that temperature and salinity changes the density of water.

Learn more about salinity here: brainly.com/question/20283396

3 0
3 years ago
Using the equation below, if you have 4.3 mol of nitrogen tribromide and
ankoles [38]

Answer:

sodium hydroxide is the limiting reactant

Explanation:

The first step is usually to put down the balanced reaction equation. This is the first thing to do when solving any problem related to stoichiometry. The balanced reaction equation serves as a guide during the solution.

2NBr3 + 3NaOH = N2 + 3NaBr + 3HOBr

Let us pick nitrogen gas as our product of interest. Any of the reactants that gives a lower number of moles of nitrogen gas is the limiting reactant.

For nitrogen tribromide

From the balanced reaction equation;

2 moles of nitrogen tribromide yields 1 mole of nitrogen gas

4.3 moles of nitrogen tribromide will yield 4.3 ×1/ 2 = 2.15 moles of nitrogen gas

For sodium hydroxide;

3 moles of sodium hydroxide yields 1 mole of nitrogen gas

5.9 moles of sodium hydroxide yields 5.9 × 1/ 3= 1.97 moles of nitrogen gas

Therefore, sodium hydroxide is the limiting reactant.

8 0
3 years ago
g, Assuming the precipitate is totally insoluble in water, which aqueous ions will be present in the solution (collected in the
Allushta [10]

Answer:

Cl⁻, Na⁺, OH⁻

Explanation:

The titration is:

CuCl₂(aq) + 2 NaOH(aq) → Cu(OH)₂(s) + 2 NaCl(aq)

In solution, before the reaction, the ions are Cu²⁺ and Cl⁻. The addition of NaOH (Na⁺ + OH⁻) produce the precipitation of Cu²⁺ forming Cu(OH)₂(s). When you reach the equivalence point, there is no Cu²⁺ because precipitates completely. All OH⁻ ions reacts when are added but when Cu²⁺ is finished, excess OH⁻ ions still in solution helping to detect the equivalence point.

Thus, ions present after the equivalence point are:<em> Cl⁻, Na⁺</em> (Don't react, spectator ions), and <em>OH⁻</em>.

3 0
3 years ago
Use the half-reaction method to balance the equation for the conversion of ethanol to acetic acid in acid solution:CH₃CH₂OH + Cr
garik1379 [7]

Balanced chemical equation is 3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O.

<h3>What is Balanced Chemical Equation ?</h3>

The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.

The chemical equation

CH₃CH₂OH + Cr₂O₇²⁻ → CH₃COOH + Cr³⁺

First assign the oxidation number for each atom in the equation.

\overset{+3}{C}\overset{+1}{H_3} \overset{-1}{C} \overset{+1}{H_2} \overset{-2}{O} \overset{+1}{H} + \overset{+6}{Cr_2} \overset{-2}{O_7} + \overset{+1}{H} \rightarrow \overset{+3}{C}\overset{+1}{H_3} \overset{+3}{C}\overset{-2}{O} \overset{-2}{O} \overset{+1}{H} + \overset{+3}{Cr}

Oxidation: C₂H₆O → C₂H₄O₂ + 4e⁻

Reduction: Cr₂O₇ + 6e⁻  → 2Cr⁺³

Now, balance the charge

Oxidation: C₂H₆O → C₂H₄O₂ + 4e⁻ + 4H⁺

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³

Now balance the oxygen atoms

Oxidation: C₂H₆O + H₂O → C₂H₄O₂ + 4e⁻ + 4H⁺

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³ + 7H₂O

Now, make electron gain equivalent to lost

Oxidation: C₂H₆O + H₂O → C₂H₄O₂ + 4e⁻ + 4H⁺ }  × 3

Reduction: Cr₂O₇ + 6e⁻ + 14H⁺ → 2Cr⁺³ + 7H₂O } × 2

Now,

Oxidation: 3C₂H₆O + 3H₂O → 3C₂H₄O₂ + 12e⁻ + 12H⁺

Reduction: 2Cr₂O₇ + 12e⁻ + 28H⁺ → 4Cr⁺³ + 14H₂O

Now, add the both equations

3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O

Thus from the above conclusion we can say that the balanced chemical equation is 3CH₃CH₂OH + 2Cr₂O₇²⁻ + 16H⁺ → 3CH₃COOH + 4Cr³⁺ + 11H₂O.

Learn more about the Balanced chemical equation here: brainly.com/question/26694427

#SPJ4

7 0
1 year ago
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